描述
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
分析
我们在Rotate List中已经讨论过这个方法，用两指针，时间O(n)，空间O(1)。
代码
Python
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class (object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
dummy = ListNode(-1)
dummy.next = head
p1 = p2 = dummy
for i in xrange(n):
p1 = p1.next
while p1.next:
p1 = p1.next
p2 = p2.next
p2.next = p2.next.next
return dummy.next
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